Python | Check if a given string is binary string or not
Last Updated :
03 May, 2023
Given string str. The task is to check whether it is a binary string or not.
Examples:
Input: str = "01010101010"
Output: Yes
Input: str = "geeks101"
Output: No
Approach 1: Using Set
- Insert the given string in a set
- Check if the set characters consist of 1 and/or 0 only.
Example:
Python3
def check(string):
p = set(string)
s = {'0', '1'}
if s == p or p == {'0'} or p == {'1'}:
print("Yes")
else:
print("No")
if __name__ == "__main__":
string = "101010000111"
check(string)
|
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
Approach 2: Simple Iteration
- Iterate for every character and check if the character is 0 or 1.
- If it is not then set a counter and break.
- After iteration check whether the counter is set or not.
Python3
def check2(string):
t = '01'
count = 0
for char in string:
if char not in t:
count = 1
break
else:
pass
if count:
print("No")
else:
print("Yes")
if __name__ == "__main__":
string = "001021010001010"
check2(string)
|
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
- Compile a regular expression using compile() for “character is not 0 or 1”.
- Use re.findall() to fetch the strings satisfying the above regular expression.
- Print output based on the result.
Python3
import re
sampleInput = "1001010"
c = re.compile('[^01]')
if(len(c.findall(sampleInput))):
print("No")
else:
print("Yes")
|
Time complexity: O(n)
Auxiliary space: O(1)
Approach 4: Using exception handling and int
Python has a built in method to convert a string of a specific base to a decimal integer, using int(string, base). If the string passed as an argument is not of the specified base, then a ValueError is raised.
Python3
def check(string):
try:
int(string, 2)
except ValueError:
return "No"
return "Yes"
if __name__ == "__main__":
string1 = "101011000111"
string2 = "201000001"
print(check(string1))
print(check(string2))
|
Time Complexity: O(1),
Auxiliary Space: O(1)
Approach 5 : Using count() function
Python3
string = "01010101010"
if(string.count('0')+string.count('1')==len(string)):
print("Yes")
else:
print("No")
|
Time Complexity: O(n), where n is number of characters in string.
Auxiliary Space: O(1)
Approach 6 : Using replace() and len() methods
Python3
string = "01010121010"
binary="01"
for i in binary:
string=string.replace(i,"")
if(len(string)==0):
print("Yes")
else:
print("No")
|
Approach 7: Using all()
The all() method in Python3 can be used to evaluate if all the letters in the string are 0 or 1.
Python3
def check(string):
if all((letter in "01") for letter in string):
return "Yes"
return "No"
if __name__ == "__main__":
string1 = "101011000111"
string2 = "201000001"
print(check(string1))
print(check(string2))
|
Time Complexity: O(n), Here n is the length of the string.
Auxiliary Space: O(1), As constant extra space is used.
Approach 8: Using issubset() method
We can use a issubset method in comprehension to check if all characters in the string are either 0 or 1.
Python3
def is_binary_string(s):
unique_chars = {c for c in s}
return unique_chars.issubset({'0', '1'})
if __name__ == "__main__":
string1 = "101011000111"
string2 = "201000001"
print(is_binary_string(string1))
print(is_binary_string(string2))
|
Time complexity: O(n), where n is the length of the input string, and
Auxiliary space: O(1).
Method: Using re.search()
- Import the re module.
- Define a regular expression that matches any character other than ‘0’ and ‘1’.
- Use the re.search() method to search for the regular expression in the given string.
- If a match is found, then the given string is not a binary string. Otherwise, it is.
Python3
import re
def is_binary_string(str):
regex = r"[^01]"
if re.search(regex, str):
return False
else:
return True
print(is_binary_string("01010101010"))
print(is_binary_string("geeks101"))
|
Time Complexity: O(n), where n is the length of the given string (as we are iterating over the string once).
Auxiliary Space: O(1)
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